Question

In the following exercises, solve the systems of equations by elimination.

$\left\{\begin{array}{l}3x+2y=-3\\ -x-2y=-19\end{array}\right.$

Short answer

The solution of the system of equations is $\left(-11,15\right)$.

Step-by-step solution

Step 1. Given Information 

We are given a system of equations.

$\left\{\begin{array}{l}3x+2y=-3\\ -x-2y=-19\end{array}\right.$

We need to solve the system using the method of elimination.

Step 2. Add the two equations

The two equations in the system are written in standard form and the coefficients of $y$are already opposites. So we add the two equations

$\begin{array}{rcl}3x+2y+(-x-2y)& =& -3-19\\ 3x+2y-x-2y& =& -22\\ 2x& =& -22\\ x& =& -\frac{22}{2}\\ x& =& -11\end{array}$

Step 3. Find the value of y

Substitute $-11$for $x$in the second equation $-x-2y=-19$.

$\begin{array}{rcl}-(-11)-2y& =& -19\\ 11-2y& =& -19\\ -2y& =& -19-11\\ -2y& =& -30\\ y& =& \frac{-30}{-2}\\ y& =& 15\end{array}$

So the solution for the given system is $(-11,15)$.

Step 4. Check the solution in the first equation

Substitute $-11$for $x$and $15$for $y$in the first equation.

$\begin{array}{rcl}3x+2y& =& -3\\ 3(-11)+2\left(15\right)& =& -3\\ -33+30& =& -3\\ -3& =& -3\end{array}$

As the statement is correct, so the solution satisfies the first equation.

Step 5. Check the solution in the second equation

Substitute $-11$for $x$and $15$for $y$in the second equation.

$\begin{array}{rcl}-x-2y& =& -19\\ -(-11)-2\left(15\right)& =& -19\\ 11-30& =& -19\\ -19& =& -19\end{array}$

As the statement is correct, so the solution satisfies the second equation.